1) 5 + (-4) = 1

2) (-8) + 2 = -6

3) 8 + (-2) = 6

4) 11 + (-3) = 8

5) (-11) + 2 = -9

6) (-7) + 3 = -4

7) (-5) + 5 = 0

8) 11 + (-12) = -1

9) (-18) + 20 = 2

10) (15) + (-12) = 3

11) (-17) + 17 = 0

12) 16 + (-2) = 14

13) (30) + (-14) = 16

14) (-19) + 20 = 1

15) (-18) + 15 = -3

16) (10) + (-6) = 4

17) (-28) + 14 = -14

18) 15 + (-30) = -15

19) (15) + (-4) = 11

20) (-21) + 11 = -10

21) 8 + (-22) = -14

22) (-15) + 4 = -11

23) (-3) + 2 = -1

24) 17 + (-14) = 3

25) 17 + (-14) = 3

thế này đọc đc mới đỉnh

\(=\sqrt{3}-5+\sqrt{3}+5=2\sqrt{3}\)

Ta có: \(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

\(=3-\sqrt{5}+3+\sqrt{5}+2\cdot\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(=6+2\cdot\sqrt{4}\)

=10

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

bai EZ​ qua

mà bài dài quá sao làm hết dc

Ta có: \(\left(2x+3\right)^2+\left(2x+5\right)^2-2\left(2x+3\right)\left(2x+5\right)\)

\(=\left(2x+3-2x-5\right)^2\)

\(=\left(-2\right)^2=4\)

Ta có:  ( - 2 ) 5 : ( - 2 ) 3 = ( - 2 ) 5 - 3 = ( - 2 ) 2 = 4 .

\(=\dfrac{3}{5}+\dfrac{2}{5}\cdot\dfrac{5}{6}=\dfrac{3}{5}+\dfrac{1}{3}=\dfrac{9}{15}+\dfrac{5}{15}=\dfrac{14}{15}\)

\(\dfrac{3}{5}+\dfrac{2}{5}.\dfrac{15}{18}=\dfrac{3}{5}+\dfrac{1}{3}=\dfrac{9}{15}+\dfrac{5}{15}=\dfrac{14}{15}\)

\(\dfrac{3}{5}+\dfrac{1}{3}=\dfrac{14}{15}\)

\(3\left(x+1\right)^2+2\left(x-3\right)^3-5\left(x-5\right)\left(x+5\right)\)

\(=3\left(x^2+2x+1\right)+2\left(x^2-6x+9\right)-5\left(x^2-25\right)\)

\(=3x^2+6x+3+2x^2-12x+18-5x^2+125\)

\(=-6x+146\)